1. Field of the Invention
This invention relates in general to vertical-scanning interferometry (VSI) for surface characterization. In particular, it relates to an error-correction method for the high-definition vertical-scan interferometric (HDVSI) procedure described in Ser. No. 11/473,447, herein incorporated by reference.
2. Description of the Related Art
The well known techniques generally classified as phase-shifting interferometry (PSI) and vertical scanning interferometry (VSI) make it possible to measure the profile of most samples. However, they do not allow low-noise measurement of samples that combine smooth surfaces with large profile gradients and discontinuities. Measuring the profile of such samples requires the large scanning range of VSI (with vertical resolution in the order of 1/100 of a wavelength), while characterizing a smooth surface texture requires the resolution normally afforded by PSI (in the order of 1/1000 of a wavelength or less).
This problem has been addressed by the development of enhanced VSI algorithms (named EVSI in the art) that combine both PSI and VSI. In particular, Ser. No. 11/473,447 disclosed a new approach (named HDVSI, from high-definition VSI) based on performing a coarse calculation of surface profile with a conventional VSI method such as center of mass (COM), quadrature center of mass (QCOM), or zero crossing detection (ZCD), and concurrently performing a phase calculation carried out using the same irradiance frame data acquired for VSI. However, the phase calculation does not utilize the conventional n-frame phase shifting approach of the prior art. Instead, it utilizes a quadrature-demodulation (QD) algorithm applied to the irradiance data contained in the VSI correlogram.
As a result of this concurrent procedure, the phase calculation is independent of the position of any particular interferometric fringe and, therefore, it is more accurate and its results are more certain than those produced by the combined VSI/PSI methods of the prior art. Once both calculations are accomplished, the phase data are incorporated into the coarse profile data through a unique “unwrapping” method that yields a final surface map with sub-nanometer resolution within a large z-height range.
The HDVSI algorithm decouples the calculation of phase from the calculation of fringe-intensity peak. This results from the fact that QD manipulation of the correlogram data does not require a prior determination of the position of the coherence intensity peak. Therefore, the calculated phase map of the surface does not inherit the errors produced by the calculation of the coarse map. Such errors are removed from the coarse map, prior to combining it with the phase map, by consistently rounding all VSI measurements to an integer multiple of 2π. In other words, the VSI coarse map is rounded to an integer multiple of λ/2. (Typically, the accuracy of a VSI coarse map is much better than λ/2.) As a result, the VSI errors are completely removed from the VSI coarse map through the rounding process, an achievement that is not possible with prior-art approaches. Thus, the HDVSI algorithm allows profiling to sub-nanometer accuracy with conventional equipment, the only necessary changes being in the computational components required to round the VSI data, to implement the QD algorithm, and to combine the phase data with the corrected VSI data in real time, as data are acquired during the scan.
Ideally, the scanner steps and the effective wavelength of the light source are constant and can be determined through precise calibration procedures. However, in reality the scanner position suffers from system errors, such as scanner nonlinearity and random deviations due to mechanical vibrations. Similarly, the effective wavelength of the light source changes with the slope of the sample surface and also as a result of scanner nonlinearities. Therefore, the scanner sampling step and the effective wavelength are not constant and cause errors that distort the measured sample-surface profile.
An analysis of the error produced by scanner position shows that, when constant filter parameters (k0 and ΔΦ0, defined below) are used and when scanner errors (δz) are present in the scanner position, the phase obtained from the HDVSI demodulation filter is a combination of the true sample phase (Φ) and a phase error (ζΦ) related to the scanner-position error. In particular, as the VSI scan is implemented through a plurality of scanning steps (ΔΦ or Δz, with reference to phase or distance, respectively) of predetermined nominal size (referred to herein interchangeably as nominal scan steps or phase steps), each detector pixel produces a signal with an amplitude that passes through and drops off very rapidly from its maximum value [Imax, corresponding to zero optical path difference (OPD)]. The interferometric signal can be expressed asI(n)=G(zn−z0)cos(2nk0Δzn−Φ),  (1)where G(zn−z0) is, for example, the Gaussian envelope of the correlogram (i.e., the amplitude of irradiance at each scanner position); z0 and zn are the positions of the scanning objective at zero OPD and at the nth scanning step, respectively; k0 is the wave number (i.e., k0=2π/λ0, λ0 being the effective wavelength of the illumination); and Δzn and Φ are the nth scan step and the phase of the correlogram, respectively. Note that the factor of 2 in the cosine function results from the fact that every scanning step Δzn produces twice that change in OPD.
As taught in Ser. No. 11/473,447, the interferometric signals I(n) are manipulated with the HDVSI algorithm to produce a low-noise sub-nanometer resolution surface map. The HDVSI algorithm comprises several steps. After undergoing pre-processing filtration, if needed, the input to the algorithm (i.e., the interferometric irradiance signals produced by the VSI scan) is processed simultaneously in two ways that allow for the independent calculation of a coarse height z0 of each surface point and of the phase Φ of the correlogram corresponding to that point. In the aggregate, the values of z0 and Φ for all surface points imaged onto the light detector yield a coarse surface map, zVSI(x,y), and a high-definition phase map, Φ(x,y), of the test surface, respectively. As one skilled in the art would readily understand, the phase map Φ(x,y) can be equivalently expressed as a fine surface (or height) map, zHD(x,y), because the phase difference between two surface points is directly proportional to the difference in their heights. Therefore, for the purposes of this disclosure, the maps Φ(x,y) and zHD(x,y) and the corresponding terms “phase map,” “surface map,” “height map” and “map” may be used interchangeably. As a separate step of the HDVSI algorithm, the two maps are judiciously combined to generate the final high-definition surface map, zHDVSI(x,y).
The phase Φ of the correlogram is determined by applying a quadrature-demodulation approach. (See, for instance, M. Fleischer et al., “Fast algorithms for data reduction in modern optical three-dimensional measurement systems with MMX technology,” Applied Optics, Vol. 39, No. 8, March 2000, pp. 1290-1297.) The signal I(n) is passed through a QD filter that performs several operations. First, the irradiance signal is differentiated to eliminate the DC component in the original correlogram. Then, it is split into “in-phase” and “quadrature” signal components, J and Q, respectively, whereinQn=G(zn−z0)cos(n2k0Δzn−Φ)sin(n2k0Δz0) and  (2a)Jn=G(zn−z0)cos(n2k0Δzn−Φ)cos(n2k0Δz0)  (2b)where Δzn is the actual size of the nth scan step and Δz0 is the parameter of the filter, chosen to match the nominal step size of the scan [typically equal to mλ0/8, m=0, 1, 2 . . . ].
The deviations of the scanner from its nominal performance may be represented by a generalized step error 5, as follows,Δzn=Δz0+δz.  (3)Assuming a nominal scan step of λ0/8 (i.e., 90°) and substituting for Δzn, Δz0 and k0 in Equations 2a and 2b, yields the following:Qn=G(n)cos [n(π/2+2k0δz)−Φ] sin(nπ/2) and  (4a)Jn=G(n)cos [n(π/2+2k0δz)−Φ] cos(nπ/2).  (4b)The quadrature demodulation procedure involves calculating the ratio
                              (                                                    ∑                                  n                  =                  1                                N                            ⁢                                                          ⁢                              Q                n                                                                    ∑                                  n                  =                  1                                N                            ⁢                                                          ⁢                              J                n                                              )                .                            (        5        )            Accordingly, from Equation 4a,
                                          ∑                          n              =              1                        N                    ⁢                                          ⁢                      Q            n                          =                              ∑                          n              =              1                        N                    ⁢                                          ⁢                                    G              ⁡                              (                n                )                                      ⁢                          cos              ⁡                              (                                                      n                    ⁡                                          (                                                                        π                          2                                                +                                                  2                          ⁢                                                      k                            0                                                    ⁢                                                      δ                            z                                                                                              )                                                        -                  Φ                                )                                      ⁢                                          sin                ⁡                                  (                                      n                    ⁢                                          π                      2                                                        )                                            .                                                          (                  6          ⁢          a                )            Equation 2b similarly yields:
                                          ∑                          n              =              1                        N                    ⁢                                          ⁢                      J            n                          =                              ∑                          n              =              1                        N                    ⁢                                          ⁢                                    G              ⁡                              (                n                )                                      ⁢                          cos              ⁡                              (                                                      n                    ⁡                                          (                                                                        π                          2                                                +                                                  2                          ⁢                                                      k                            0                                                    ⁢                                                      δ                            z                                                                                              )                                                        -                  Φ                                )                                      ⁢                                          cos                ⁡                                  (                                      n                    ⁢                                          π                      2                                                        )                                            .                                                                    (                      6            ⁢            b                    )                )            
Applying to Equation 6a the general trigonometric formulassin(a+b)=sin(a)cos(b)+cos(a)sin(b) and  (7a)sin(a−b)=sin(a)cos(b)−cos(a)sin(b),  (7b)which, combined, produce the relationsin(a+b)−sin(a−b)=2 cos(a)sin(b),  (8)Equation 6a becomes
                              2          ⁢                                    ∑                              n                =                1                            N                        ⁢                                                  ⁢                          Q              n                                      =                                            ∑                              n                =                1                            N                        ⁢                                                  ⁢                                          G                ⁡                                  (                  n                  )                                            ⁢                              sin                ⁡                                  (                                                            n                      ⁢                                                                                          ⁢                      π                                        +                                          2                      ⁢                                              nk                        0                                            ⁢                                              δ                        z                                                              -                    Φ                                    )                                                              -                                    ∑                              n                =                1                            N                        ⁢                                                  ⁢                                          G                ⁡                                  (                  n                  )                                            ⁢                              sin                ⁡                                  (                                                            2                      ⁢                                                                                          ⁢                                              nk                        0                                            ⁢                                              δ                        z                                                              -                    Φ                                    )                                                                                        (        9        )            The first summation term is approximately zero when N>>1; therefore,
                              2          ⁢                                    ∑                              n                =                1                            N                        ⁢                                                  ⁢                          Q              n                                      =                              -                                          ∑                                  n                  =                  1                                N                            ⁢                                                          ⁢                                                G                  ⁡                                      (                    n                    )                                                  ⁢                                  sin                  ⁡                                      (                                                                  2                        ⁢                                                                                                  ⁢                                                  nk                          0                                                ⁢                                                  δ                          z                                                                    -                      Φ                                        )                                                                                =                                                    -                                  cos                  ⁡                                      (                    Φ                    )                                                              ⁢                                                ∑                                      n                    =                    1                                    N                                ⁢                                                                  ⁢                                                      G                    ⁡                                          (                      n                      )                                                        ⁢                                      sin                    ⁡                                          (                                              2                        ⁢                                                  nk                          0                                                ⁢                                                  δ                          z                                                                    )                                                                                            +                                          sin                ⁡                                  (                  Φ                  )                                            ⁢                                                ∑                                      n                    =                    1                                    N                                ⁢                                                                  ⁢                                                      G                    ⁡                                          (                      n                      )                                                        ⁢                                      cos                    ⁡                                          (                                              2                        ⁢                                                  nk                          0                                                ⁢                                                  δ                          z                                                                    )                                                                                                                              (                  10          ⁢                                          ⁢          a                )            which follows from Equation 7b. Similarly,
                              2          ⁢                                    ∑                              n                =                1                            N                        ⁢                                                  ⁢                          J              n                                      =                                            cos              ⁡                              (                Φ                )                                      ⁢                                          ∑                                  n                  =                  1                                N                            ⁢                                                          ⁢                                                G                  ⁡                                      (                    n                    )                                                  ⁢                                  cos                  ⁡                                      (                                          2                      ⁢                                                                                          ⁢                                              nk                        0                                            ⁢                                              δ                        z                                                              )                                                                                +                                    sin              ⁡                              (                Φ                )                                      ⁢                                          ∑                                  n                  =                  1                                N                            ⁢                                                          ⁢                                                G                  ⁡                                      (                    n                    )                                                  ⁢                                  sin                  ⁡                                      (                                          2                      ⁢                                                                                          ⁢                                              nk                        0                                            ⁢                                              δ                        z                                                              )                                                                                                          (                  10          ⁢                                          ⁢          b                )            
Equations 10a and 10b may be further simplified by setting
                                          ∑                          n              =              1                        N                    ⁢                                          ⁢                                    G              ⁡                              (                n                )                                      ⁢                          cos              ⁡                              (                                  2                  ⁢                                                                          ⁢                                      nk                    0                                    ⁢                                      δ                    z                                                  )                                                    =                  a          ⁢                                          ⁢          and                                    (                  11          ⁢                                          ⁢          a                )                                                      ∑                          n              =              1                        N                    ⁢                                          ⁢                                    G              ⁡                              (                n                )                                      ⁢                          sin              ⁡                              (                                  2                  ⁢                                                                          ⁢                                      nk                    0                                    ⁢                                      δ                    z                                                  )                                                    =        b                            (                  11          ⁢                                          ⁢          b                )            which yields
                                          2            ⁢                                          ∑                                  n                  =                  1                                N                            ⁢                                                          ⁢                              Q                n                                              =                                                    -                                  cos                  ⁡                                      (                    Φ                    )                                                              ⁢              b                        +                                          sin                ⁡                                  (                  Φ                  )                                            ⁢              a                                      ,                                  ⁢        and                            (                  12          ⁢                                          ⁢          a                )                                          2          ⁢                                    ∑                              n                =                1                            N                        ⁢                          J              n                                      =                                            cos              ⁡                              (                Φ                )                                      ⁢            a                    ⁢                                          +                                    sin              ⁡                              (                Φ                )                                      ⁢                          b              .                                                          (                  12          ⁢                                          ⁢          b                )            Defining the error in terms of a new variable ζz in phase space, such that
                              cos          ⁡                      (                          ζ              z                        )                          =                              a                                                            a                  2                                +                                  b                  2                                                              ⁢                                          ⁢          and                                    (                  13          ⁢                                          ⁢          a                )            
                                          sin            ⁡                          (                              ζ                z                            )                                =                      b                                                            a                  2                                +                                  b                  2                                                                    ,                            (                  13          ⁢                                          ⁢          b                )            and further dividing Equations 12a and 12b by √{square root over (a2+b2)} and substituting Equations 13a and 13b produces the following relations in terms of the phase error ζz:
                                                                        2                ⁢                                                      ∑                                          n                      =                      1                                        N                                    ⁢                                                                          ⁢                                                            Q                      n                                        /                                                                                            a                          2                                                +                                                  b                          2                                                                                                                                =                                                                    -                                          cos                      ⁡                                              (                        Φ                        )                                                                              ⁢                                      sin                    ⁡                                          (                                              ζ                        z                                            )                                                                      +                                                      sin                    ⁡                                          (                      Φ                      )                                                        ⁢                                      cos                    ⁡                                          (                                              ζ                        z                                            )                                                                                                                                                              =                                  sin                  ⁢                                      (                                          Φ                      -                                              ζ                        z                                                              )                                                              ,              and                                                          (                  14          ⁢          a                )                                                                                    2                ⁢                                                      ∑                                          n                      =                      1                                        N                                    ⁢                                                                          ⁢                                                            J                      n                                        /                                                                                            a                          2                                                +                                                  b                          2                                                                                                                                =                                                                    cos                    ⁡                                          (                      Φ                      )                                                        ⁢                                      cos                    ⁡                                          (                                              ζ                        z                                            )                                                                      +                                                      sin                    ⁡                                          (                      Φ                      )                                                        ⁢                                      sin                    ⁡                                          (                                              ζ                        z                                            )                                                                                                                                              =                                                cos                  ⁡                                      (                                          Φ                      -                                              ζ                        z                                                              )                                                  .                                                                        (                  14          ⁢          b                )            
Finally, dividing Equations 14a and 14b yields
                                          (                                                            ∑                                      n                    =                    1                                    N                                ⁢                                                                  ⁢                                  Q                  n                                                                              ∑                                      n                    =                    1                                    N                                ⁢                                                                  ⁢                                  J                  n                                                      )                    =                                    sin              ⁡                              (                                  Φ                  -                                      ζ                    z                                                  )                                                    cos              ⁡                              (                                  Φ                  -                                      ζ                    z                                                  )                                                    ,                                            or              ⁢                                                          ⁢              Φ                        -                          ζ              z                                =                                                    tan                                  -                  1                                            ⁡                              (                                                                            ∑                                              n                        =                        1                                            N                                        ⁢                                                                                  ⁢                                          Q                      n                                                                                                  ∑                                              n                        =                        1                                            N                                        ⁢                                                                                  ⁢                                          J                      n                                                                      )                                      .                                              (        15        )            From Equation 15 it is apparent that, when constant filter parameters (k0,Δzn) are used and scanner errors (ζz) are present during the scan, the phase obtained from the demodulation filter is the combination of the true sample phase Φ and a phase error ζz related to the scanner-position error.
If the scanner-position error is random, then the quadrature demodulation process automatically suppresses the error. Referring to Equation 11b, it is clear that a small random error δz will produce positive and negative values that cancel out, that is,
                                          ∑                          n              =              1                        N                    ⁢                                          ⁢                                    G              ⁡                              (                n                )                                      ⁢                          sin              ⁡                              (                                  2                  ⁢                                      nk                    0                                    ⁢                                      δ                    z                                                  )                                                    =                  b          ≈          0.                                    (        16        )            Accordingly,
                              cos          ⁡                      (                          ζ              z                        )                          =                              a                                                            a                  2                                +                                  b                  2                                                              ≅                      1            ⁢                                                  ⁢            and                                              (                  17          ⁢                                          ⁢          a                )                                                      sin            ⁡                          (                              ζ                z                            )                                =                                    b                                                                    a                    2                                    +                                      b                    2                                                                        ≅            0                          ,                            (                  17          ⁢                                          ⁢          b                )            which means that ζz=0 and, in turn,
                                          tan                          -              1                                ⁡                      (                                                            ∑                                      n                    =                    1                                    N                                ⁢                                                                  ⁢                                  Q                  n                                                                              ∑                                      n                    =                    1                                    N                                ⁢                                                                  ⁢                                  J                  n                                                      )                          =                              Φ            -                          ζ              z                                =                      Φ            .                                              (        18        )            This is demonstrated in FIG. 1, wherein the fringe-like noise produced by random scanner-position error on phase calculated using conventional n-frame PSI algorithms is illustrated in the top image of the figure. As seen in the bottom image, the integration operation in the HDVSI demodulation filter causes the error term δz to near zero, so the fringe noise is effectively suppressed.
Thus, the remaining scanner-position errors are produced by scanner nonlinearity. This invention is directed at providing real-time correction of such errors. A related approach for the correction of errors caused by changes in the system's effective wavelength is described in a separate disclosure.